| New Reply |
Definition of relative error |
Share Thread | Thread Tools |
| Sep16-12, 04:49 PM | #1 |
|
|
Definition of relative error
Hi all,
I have a general question about relative error. Suppose that we have a vector of measurements [itex]\hat{b}=\left(\hat{b_{1}},\hat{b_{2}},...,\hat{b_{n}}\right)[/itex]. Furthermore, suppose that these measurements are accurate to 10%. My natural interpretation of this statement is that there is a "true" vector [itex]b=\left(b_{1},b_{2},...,b_{n}\right)[/itex] such that [itex]\frac{\left|b_{1}-\hat{b_{1}}\right|}{\left|b_{1}\right|}[/itex], [itex]\frac{\left|b_{2}-\hat{b_{2}}\right|}{\left|b_{2}\right|}[/itex], ..., [itex]\frac{\left|b_{n}-\hat{b_{n}}\right|}{\left|b_{n}\right|}≤0.1[/itex]. I have seen in the literature that we can use the maximum norm of a vector to define the relative error. So, the relative error in [itex]\hat{b}[/itex] could be defined as [itex]\frac{\left\|b-\hat{b}\right\|}{\left\|b\right\|}[/itex] where [itex]\left\|v\right\|=\max\limits_{i} \left|v_i\right|[/itex]. The problem that I find with this is the fact that we can't conclude anything about the individual entries from this definition. For example, if [itex]b=\left(1,2,3\right)[/itex] and [itex]\hat{b}=\left(1.14,1.9,3.15\right)[/itex], then [itex]\frac{\left\|b-\hat{b}\right\|}{\left\|b\right\|}=\frac{0.15}{3}=0.05≤0.1[/itex] which indicates that the relative error in [itex]\hat{b}[/itex] is less than 10%. On the other hand, the relative error in the first entry of [itex]\hat{b}[/itex] is [itex]\frac{0.14}{1}=0.14≥0.1[/itex]. Now, suppose we solve the systems [itex]A\hat{x}=\hat{b}[/itex] and [itex]Ax=b[/itex] where [itex]A[/itex] is invertible. According to the literature, [itex]\frac{\left\|\hat{x}-x\right\|}{\left\|x\right\|}≤\left\|A^{-1}\right\|\left\|A\right\|\frac{\left\|\hat{b}-b\right\|}{\left\|b\right\|}[/itex] Where the norm of a matrix [itex]A[/itex] is defined to be [itex]\max\limits_{i} \sum\limits_{j} \left|a_{ij}\right|[/itex]. If we know that the relative error in [itex]\hat{b}[/itex] is less than 10%, then we can put a bound on the relative error in [itex]\hat{x}[/itex]: [itex]\frac{\left\|\hat{x}-x\right\|}{\left\|x\right\|}≤0.1\left\|A^{-1}\right\|\left\|A\right\|[/itex] But as shown above, this does not put a bound on the relative error in the individual entries of [itex]\hat{x}[/itex]. So my question is, what is the point of finding the relative error in the vector if we cannot use that to put a bound on the relative error of the individual entries? Maybe I'm misinterpreting something here? Thanks! |
| New Reply |
| Tags |
| error bound, norm, relative error |
| Thread Tools | |
Similar Threads for: Definition of relative error
|
||||
| Thread | Forum | Replies | ||
| Relative error definition of entropy | Advanced Physics Homework | 0 | ||
| Thermo. Relative Error | Introductory Physics Homework | 0 | ||
| Relative Error!? | Introductory Physics Homework | 8 | ||
| relative error from heck | Introductory Physics Homework | 0 | ||
| absolute and relative error | Introductory Physics Homework | 1 | ||